Solve for Int: cos(x)=cos(17π/5)

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Homework Help Overview

The discussion revolves around solving the equation cos(x) = cos(17π/5), focusing on the periodic nature of the cosine function and the implications for determining values of x.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants explore the periodicity of the cosine function and how it relates to the given angle. There are questions regarding the substitution of integer values and the reasoning behind arriving at specific angles like 7π/5.

Discussion Status

Some participants have provided insights into the periodic properties of cosine, suggesting that adding multiples of 2π yields equivalent angles. However, there is still some confusion regarding the notation used and the interpretation of certain expressions.

Contextual Notes

There are indications of potential typos in the expressions used, and the clarity of the original poster's statements is questioned. The discussion also touches on the need for further clarification regarding the integer values involved in the solution process.

Karma
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sin-(cos(17pi/5))

cos=(x+n2pi)=cos(x)

I know that this becomes 7pi/5.. But what is substituted for the integer value? and why? (how does it become 7pi/5)
 
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because cosinus is periodic with period 2pi. So if you add 2pi in the argument you get the same result.
 
Karma said:
sin-(cos(17pi/5))

cos=(x+n2pi)=cos(x)
It's not at all clear what you mean by "sin-(cos(17pi/5)". do you mean 'sine or cosine of that'? And, of course the "=" in "cos=" is a typo.

17/5= 3 and 2/5= 2+ (1+ 2/5). The "n" in "n2pi" is 1 and the "x" is (1+ 2/5)pi= (7pi/5).
 
Karma said:
sin-(cos(17pi/5))

cos=(x+n2pi)=cos(x)
It's not at all clear what you mean by "sin-(cos(17pi/5)". do you mean 'sine or cosine of that'? And, of course the "=" in "cos=" is a typo.

17/5= 3+ 2/5= 2+ (1+ 2/5). The "n" in "n2pi" is 1 and the "x" is (1+ 2/5)pi= (7pi/5).
 

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